Differentiation Rules, Implicit Differentiation, and Related Rates
The algebra of derivatives - product, quotient and chain rules - plus differentiating relations implicitly and linking rates that move together.
Leads to: The chain rule reappears as Itô’s lemma (Phase 9); implicit differentiation underlies option greeks and yield curves.
Learning Objectives
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- Apply the power, product, quotient, and chain rules to differentiate compositions.
- Differentiate implicitly a relation that does not isolate y and solve for dy/dx.
- Set up and solve a related-rates problem by differentiating a constraint with respect to time.
- Verify a symbolic derivative numerically with a central difference.
Key Vocabulary
- Power rule
- d/dx x^n = n x^{n-1} for any real n.
- Product rule
- (fg)′ = f′g + fg′.
- Quotient rule
- (f/g)′ = (f′g − fg′)/g².
- Chain rule
- (f∘g)′(x) = f′(g(x))·g′(x); differentiate outer, keep inner, times inner′.
- Implicit differentiation
- Differentiating both sides of F(x,y)=0 treating y as a function of x, then solving for y′.
- Related rates
- Using a constraint to relate the time-derivatives of interdependent quantities.
The four rules you compose everything from
With four rules you can differentiate essentially any elementary formula. The subtle one is the chain rule: a composition’s rate is the outer rate times the inner rate.
Implicit differentiation
When a curve is given by a relation you cannot (or would rather not) solve for \(y\), differentiate both sides treating \(y=y(x)\) and using the chain rule on every \(y\) term, then solve for \(y'\).
Related rates
If two quantities satisfy a relation and both change with time, differentiating the relation with respect to \(t\) links their rates. Example: a spherical balloon with \(V=\tfrac43\pi r^3\) gives \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\), so a known inflation rate \(dV/dt\) determines how fast the radius grows.
Interactive: check a derivative numerically
- Forgetting the inner derivative in the chain rule - writing (e^{3x})′=e^{3x} instead of 3e^{3x}.
- Dropping the y′ factor when differentiating a y term implicitly (that factor is the chain rule).
- Mixing up the quotient-rule sign: the numerator is f′g − fg′, in that order.
- In related rates, plugging in specific numbers BEFORE differentiating, which freezes the variables.
- Differentiate the general relation first; substitute the instantaneous values only at the end.
- For nested compositions, peel from the outside in, multiplying each layer’s derivative.
- A central difference with h≈1e−5 is a cheap, reliable check on any hand derivative.
- Implicit differentiation gives greeks: differentiating a pricing relation yields sensitivities like delta and gamma.
Knowledge Check
Practical Exercise
A company’s revenue is \(R(q)=q\,p(q)\) where price depends on quantity via \(p(q)=100-2q\). (a) Find marginal revenue \(R'(q)\). (b) At what \(q\) is marginal revenue zero, and what does that mean?
(a) \(R(q)=q(100-2q)=100q-2q^2\), so by the power rule \(R'(q)=100-4q\). (You can also use the product rule on \(q\cdot p(q)\): \(R'=p(q)+q p'(q)=(100-2q)+q(-2)=100-4q\).)
(b) Set \(R'(q)=0\): \(100-4q=0\Rightarrow q=25\). Marginal revenue zero means selling one more unit adds no revenue - the revenue curve peaks here (a maximum, since \(R''=-4\lt 0\)). Beyond \(q=25\) the price cut outweighs the extra volume.
Lesson Summary
Formula Sheet Additions
Retrieval Practice
Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.
A: (f∘g)′(x)=f′(g(x))g′(x): differentiate the outer function at the inner value, then multiply by the inner derivative. The common error is omitting the inner derivative g′(x).
A: Write a relation among the quantities, differentiate it with respect to time (chain rule on each variable), then substitute the given instantaneous values and solve for the unknown rate - substituting numbers only after differentiating.
Completion Checklist
- I can explain the core ideas in my own words
- I worked the derivations/examples by hand
- I completed the interactive workbench(es)
- I passed the knowledge check